Solve for $z$. $12 = 15\left(z-\dfrac15\right)$
Solution: Let's divide and then add to get $z$ by itself. $\begin{aligned}12 &= 15\left(z-\dfrac15\right)\\ \\\\\\ \dfrac{12}{{15}} &= \dfrac{15\left(z-\dfrac15\right)}{{15} }~~~~~~~\text{divide each side by } {15}\\ \\\\ \dfrac{12}{{15}}&=\dfrac{\cancel{15}\left(z-\dfrac15\right)}{\cancel{{15}}} \\ \\ \dfrac{12}{{15}} &= z-\dfrac15\end{aligned}$ $\begin{aligned}\dfrac45&=z-\dfrac15\\ \\ \dfrac45 {+\dfrac15}&= z-\dfrac15{+\dfrac15}~~~~~~~{\text{add }\dfrac15} \text{ to each side} \text{ to get } z \text{ by itself }\\ \\ \dfrac45{+\dfrac15}&= z-\cancel{ \dfrac15} {{+}\cancel{{\dfrac15}}}\\ \\ \dfrac45{+\dfrac15}&=z\end{aligned}$ The answer: $z={1}$ Let's check to make sure. $\begin{aligned} 12 &= 15\left(z-\dfrac15\right) \\\\ 12 &\stackrel{?}{=}15\left({1}-\dfrac15\right) \\\\ 12&\stackrel{?}{=} 15\left(\dfrac45\right) \\\\ 12&\stackrel{?}{=} \dfrac{60}5\ \\\\ 12 &= 12 ~~~~~~~~~~~~~~\text{Yes!} \end{aligned}$